2
$\begingroup$

Let's say we have the time-independent Schrödinger Equation for a free particle (without potential): $$-\frac{\hbar}{2m} \frac{d^2 \psi}{dx^2} = E\psi$$ Whose general solution is: $$\psi(x) = Ae^{ikx} + Be^{-ikx}$$ Where $k = \frac{2\pi}{\lambda}$ (i am supposing just one dimension, hence the use of a scalar quantity). Let's say that i have the following contour values: $\psi(0) = 0$ and $\psi(L) = 0 $ (First Question: What is the meaning of this? I know that this means that there is no probability of finding the particle in either $x=0$ or $x=L$. But how does this relate physically to the system and why can i impose this? A particle with no force shouldn't be an uniform distribution over $[0,L]$? (with me not being allowed to set these contour values). To help solving the contour value, i will put the solution in real form: $$\psi(x) = A\cos(kx) + B\sin(kx)$$ Using the first condidtion: $$\psi(0) = 0 \rightarrow A\cos(0) + B\sin(0) = 0 \rightarrow \boxed{A = 0}$$ Now for the second condition: $$\psi(L) = 0 \rightarrow B\sin(kL) = 0 \rightarrow kL = \Big(n+\frac12\Big)\pi$$ $$ k = \frac{\big(n + \frac12 \big)\pi}{L}$$ Finally, the solution is: $$\psi(x) = B \sin \Big[\frac{(n+\frac12)\pi}{L}x\Big]$$ Now, the second question is: What is the meaning of $n$? If i choose, for example, $n=1$. What does that mean? Why does the potential and the contour values can't tell us how the system will evolve? I know that $n$ is analogous to the "harmonics" of a wave. But i can't understand how does that relate to a particle in QM.

$\endgroup$
3
  • $\begingroup$ If you apply this to the Hydrogen Atom, $n$ is the quantum number that characterizes the energy levels. $\endgroup$ Nov 7 '17 at 11:48
  • $\begingroup$ "Let's say that i have the following contour values: $\psi(0)=0$ and $\psi(L)=0$." - Where did you get that contour values from? That would mean that the wavefunction vanishes at those points. It's like you're trying to solve the particle in a box problem for particle being free between $x=0$ and $x=L$. $\endgroup$ Nov 7 '17 at 12:00
  • $\begingroup$ @Arthur they are the contour values my (not so good) teacher decided to use without explaining why exactly. $\endgroup$ Nov 7 '17 at 12:04
1
$\begingroup$

The difficulty with your question is that, by forcing time-independent conditions such as $\psi(0)=\psi(L)=0$, the solutions become standing waves rather than plane waves. To be explicit, you can restore full time dependence $$ \Psi(x,t)=Ae^{i(kx-\omega t)}+Be^{-i(kx+\omega t)} $$ and see that the $e^{i(kx-\omega )t}$ and $e^{-i(kx+\omega t)}$ correspond to two plane waves travelling in different directions, with momentum $p=\hbar k$ in reverse directions. Thus your real form $\psi(x)\sim \sin(kx)$ does not describe a free particle since it does not correspond to a single value of momentum. Since this solution is not for a free particle, it cannot be expected to have a uniform probability density.

In a more mathematical language, the solution in terms of sine is not an eigenfunction of the momentum operator $-i\hbar d/dx$.

Once you have standing waves you have (loosely speaking) quantization of energy values, with only a selected number of energies allowed. In your case, your standing waves must have a wavelength that is a multiple of $L/2$. This implies some conditions on $\vert k\vert$ and, since $p=\hbar k$, some conditions on $p^2$ and thus on the allowed energy.

$\endgroup$
8
  • $\begingroup$ But the probability density $|\psi(x)|^2 = \psi \bar{\psi}$ still depends on $x$, even for a general solution. Why is that? $\endgroup$ Nov 7 '17 at 12:36
  • 1
    $\begingroup$ This is a good answer but "then $\psi(x)$... does not describe a free particle since it does not correspond to a single value of momentum" is not right. Did you mean to say something slightly different? $\endgroup$
    – knzhou
    Nov 7 '17 at 12:38
  • $\begingroup$ @VitorCGoergen the plane wave solution $\psi(x)\sim e^{i kx}$ has $\vert \psi(x)\vert^2$ constant. The general solution is superposition of $\pm k$ so does not describe a plane wave until you choose a direction of propagation. $\endgroup$ Nov 7 '17 at 12:50
  • $\begingroup$ @knzhou Back. would appreciate if you could clarify your comment so I can make appropriate correction. $\endgroup$ Nov 7 '17 at 16:48
  • $\begingroup$ @ZeroTheHero why do you consider only $e^{ikx}$ when talking about $|\psi|^2$? If the general solution is $\psi(x) = Ae^{ikx} + Be^{-ikx}$, why do we "throw away" $Be^{-ikx}$? Because for the general solution the probability density won't stay constant. Besides this, why does a property (in this case, the probability density function) can have one property not depending on any initial condition (i.e, $|\psi(x)|^2 =$ constant) but when you apply an initial condition, this property changes? (for the solution i found, the probability distribution isn't uniform). $\endgroup$ Nov 7 '17 at 19:19
0
$\begingroup$

As to your first question: the meaning is that the particle is not a free particle after all. This is the problem of a particle in a one-dimensional infinitely deep square potential well: $\Phi=0$ for $x\in[0,L]$ but $\Phi=\infty$ otherwise. Then (by Schrödinger's equation) $\psi=0$ at those points where $\Phi=\infty$. Thus the particle is only free to move inside the potential well, but not outside.

As to your second question: The meaning of $n$ is that there are only a finite, countable number of independent solutions to Schrödinger's equation with $n$ the index into those solutions. $n$ is called their quantum number. If you work out the energy of each solution $$ E_n = \frac{\hbar}{2m}\left[\frac{(n+\frac{1}{2})\pi}{L}\right]^2 $$ you see that it increases with $n$, but also that $E_0>\Phi=0$, the ground-state energy. A general wave function $\psi(x)$ can be decomposed into a superposition of these discrete wave solutions (aka a Fourier series).

Your question regarding evolution requires to solve the time-dependent Schrödinger equation, which you haven't.

$\endgroup$
0
$\begingroup$

The boundary values $\psi(0) = \psi(L) = 0$ makes no sense for a free particle. For an absolutely free particle(having one degree of freedom), the only constraints are $V(x) = 0$ $ \forall x \in S$, where $S$ denotes the space in which the particle exists, and $\int _{S} \psi^*\psi = 1$ (Normalization condition for a particle). No other constraint is obvious from this problem. Hence your contours are not applicable here. The constraints you mentioned make sense if for example the particle is inside a one-dimensional infinite potential well. This means the following: $V(0) = V(L) = \infty$ and $V(x) = 0$ $\forall$ $x | 0<x<L$. Obviously the normalization applies as well. The solutions for $\psi$ can be found using trial solutions, and it gives the answer you mentioned. In this case, we can put $\psi(0) = \psi(L) = 0$ as, the particle cannot be 'at' the wall. It will bounce off it. Now, in a ground state(energy), it is expected that the particle will spend more time around the centre of the box and lesser time near the edges, as it bounces off the walls, and hence the solution is also slightly intuitive. Hope some of this helps.

$\endgroup$
-3
$\begingroup$

QM->quantum mechanics;

first question:

QM assumes that everything in existence can be described by a wavefunction. And for the power to be able to describe something physical it must be continuous, differentiable and normalizable in all of space.(that are the postulates with which QM proceeds)

now it is deducible that for the wavefunction to be continuous it must reduce to zero at the boundary so as to be continuous before and after it.

second question:

you introduces a variable to satisfy that kx = 0 at x=0 and x=L. additionally n also gifts us the result that the energy is quantized for our wavefunction.

after solving we get: E proportional to n^2

so for different values of n we get different physical properties for the wavefunction and so n earns a title of state variable and becomes important for us to consider for our wavafunction.

now we know that for same potential energy different energy states for the same wavefunction can exist and the contour values will change accordingly.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.